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已知函数f(x)=x²+2x-3,x∈[t,t+2]。求函数的值域
解析:
f(x)
=x²+2x-3
=(x+1)²-4
(1) t+2≤-1时,f(x)在[t,t+2]上单调递增减
fmax=t²+2t-3
fmin=t²+6t+5
值域:[t²+6t+5,t²+2t-3]
(2) t≥-1时,f(x)在[t,t+2]上单调递增增
fmin=t²+2t-3
fmax=t²+6t+5
值域:[t²+2t-3,t²+6t+5]
(3) t<-1<t+2时,
fmin=-4
fmax=M=max{t²+2t-3,t²+6t+5}
值域:[-4,M]
解析:
f(x)
=x²+2x-3
=(x+1)²-4
(1) t+2≤-1时,f(x)在[t,t+2]上单调递增减
fmax=t²+2t-3
fmin=t²+6t+5
值域:[t²+6t+5,t²+2t-3]
(2) t≥-1时,f(x)在[t,t+2]上单调递增增
fmin=t²+2t-3
fmax=t²+6t+5
值域:[t²+2t-3,t²+6t+5]
(3) t<-1<t+2时,
fmin=-4
fmax=M=max{t²+2t-3,t²+6t+5}
值域:[-4,M]
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